What is the Correct Form of the Free Propagator in QFT?

  • Thread starter Thread starter Phymath
  • Start date Start date
  • Tags Tags
    Propagator Qft
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
Phymath
Messages
183
Reaction score
0

Homework Statement



from Zee QFT in a nutshell

the free propagator between two "sources" on the field is given by[tex]D(x_\mu) = -i \int \frac{d^3k}{(2\pi)^3 2 \omega_k}[e^{-i(\omega_kt-k\bullet x)} \Theta(x_0) + e^{i(\omega_k t-k\bullet x)} \Theta(-x_0)[/tex]

for a space like separation ([tex]x_0 = 0[/tex]) Zee gets

[tex] -i\int\frac{d^3k}{(2\pi)^3 2 \omega_k}e^{-i k\bullet x}[/tex]

with assumption that [tex]\Theta(0) = 1/2[/tex]

with that assumption i don't agree with Zee i get

[tex] -i\int\frac{d^3k}{(2\pi)^3 2 \omega_k}cos(k \bullet x)[/tex]

where am I going wrong?
 
Physics news on Phys.org
The two expressions are equal. If you write the complex exponential as a sum of a sine and cosine, the sine term will integrate to zero because it is odd in k.
 
In the same book, in this definition of the D(x). Why do we get a term exp^-i(ωt-kx) when X_o in positive and a term exp^i(ωt-kx) when X_o is negative?